In all of your solutions,
- clearly show your system definition with proper sketches indicating related energy flows
- clearly show how you write your equations (e.g. Mass, Energy, Entropy, Exergy balance, etc.)
- identify relevant energy flows
- use of definitions of power-energy relationships and efficiencies as needed
- use of proper equation of state, or property relations as needed
- clearly show your steps, assumptions and approximations.
Microsoft Word - ENSY_5000_2022_Fall_HW_06.docx ENSY 5000 Fundamentals of Energy System Integration Homework Problem Set 6 READING ASSIGNMENT: Read Chapters 5 and 6 of the Textbook. NOTE: To solve this problem set as well as any other problem set, it will help if you draw your system and identifying the energy component (property) changes and the energy interactions at the boundaries. In some cases, you may need a sub system analysis to identify the energy interactions. In all of your solutions, clearly show your system definition with proper sketches indicating related energy flows clearly show how you write your equations (e.g. Mass, Energy, Entropy, Exergy balance, etc.) identify relevant energy flows use of definitions of power-energy relationships and efficiencies as needed use of proper equation of state, or property relations as needed clearly show your steps, assumptions, and approximations. Q1: In the previous HW set, you evaluated a transient (time dependent) process using a numerical multi‐step technique until steady‐state is reached (see figure 1 on the right). Use the same technique, and modify your solution for to evaluate entropy production and exergy destruction for the entire process. Since, temperature of the hot and cold liquids will change during the process, entropy production for each step will be different. Suppose j is the step number. At step j, temperature of hot fluid goes from TH,j TH,j+1 and temperature of the cold fluid goes from TC,j TC,j+1. FIGURE 1 We expect ? , ? , 0 will be negative since TH is decreasing and ? , ? , 0 since TC is increasing during this process. You evaluated heat passing from hot to cold liquid for each step (?? ) in the prior assignment. Having all these parameters, you can compute entropy production in hot the liquid, in the cold liquid, and in between where heat passes from one side to other using entropy balance equation for each step. Please note that small entropy changes in the hot and cold liquids, assuming incompressible substance would be Δ? , ? , ? , ? ? ln , , and Δ? , ? , ? , ? ? ln , , You should also consider incremental entropy transfer by heat term for incremental each step, for example ?? , and ?? , . Boundary temperature here will change at each step as well as during the step. However, if you keep incremental temperature changes in the hot and cold liquids small, approximation errors here will be small. You can either set Tb to the temperature at the beginning of the step or set Tb to the average of temperatures at the beginning and end of the step. For example, if you want to set it to the temperature at the beginning of the step Tb for hot side becomes ? , ? , or if you prefer to use average temperature ? , ? , ? , /2. Both of them are approximations with different approximation errors. The latter will give you smaller error than the former. If temperature change is small for each step approximation error will be small in both options. Once you have entropy change and entropy transfer by heat, you can compute small incremental entropy production for each step using entropy balance equation. Let’s call them ?? , for hot side ?? , and for cold side. If you add all incremental entropy changes of every step you can find entropy production for the entire process, i.e. ? , ∑ ?? , assuming we have n steps to reach equilibrium. You have to do all these calculations for the interface between hot liquid and cold liquid where heat passes through and temperature changes along the path, which is indicated by red dashed line in Figure 1. For the interface, assume that there is no mass, and no entropy (or no entropy change) but there is entropy transfer by heat and hence there is entropy production. Then find entropy production in the interface ? , . You can find exergy destruction directly by multiplying entropy production with the dead state temperature (? ). You can assume ? 275 K. Compare entropy production, and exergy destruction for the entire process to the values you have found in Q2 of previous assignment. Q2: Consider the conditions of Q1 above. Instead of having a direct heat from hot to cold liquids through an interface, consider the following proposal that is sketched in Figure 2. Heat is allowed to flow from hot liquid (acting as energy source) through a Carnot heat engine (power cycle) and then engine rejects heat to the cooler liquid (acting as the heat sink). Temperature of each reservoir interacting with Carnot heat engine is changing unlike ideal Carnot cycle where reservoir temperatures are constant. FIGURE2 Since the temperature of the boundary that heat is passing through (both at the hot and cold) are changing, the Carnot efficiency is also changing during the process. To simplify the solution, we can consider the entire process from initial states to final equilibrium occurs in quasi-equilibrium conditions. That is, process is completed in piece-wise steps such that at the end of each small step equilibrium is reached (Similar to Q1 above). Figure 2 shows energy flows for one step. Determine A) final equilibrium temperature of liquids inside the beakers (this will be different than result in Q1, see hint 1 below) B) Plot how the temperatures of the hot and cold liquid changes as a function of steps (or time if you prefer) until equilibrium is reached. C) the amount of total work developed by the Carnot heat engine, which would be sum of all incremental work you can find for each step, D) Plot how the incremental work produced by the engine changes until the equilibrium is reached. E) total heat taken from the warm liquid (it will be different than result in Q1), F) total heat discharged to the cold liquid (it will be different than result in Q1), G) entropy change of hot and cold liquids and compare them with results of Q1, H) entropy produced in the cold, in the warm liquid, and in the heat engine, separately. I) total entropy produced including cold and hot liquids, heat engine and everything in between if any. J) total exergy destroyed in the cold liquid and in the warm liquid separately. K) total exergy destroyed in all components including the heat engine (power cycle). L) the differences between the results of this problem and Q1 above and comment on them. HINT1: Follow step by step approach. For each step (say step j), you should determine the incremental work produced (Wnet,j) by the heat engine. You should find the incremental heat coming from hot liquid (QH,j) and incremental heat discharged to cold liquid (QC,j). QH,j and QC,j will not be equal to each other due to heat engine (they were the same in Q1). You can find resulting incremental temperature changes of both liquids (TC,j will be different than TH,j due to mass and heat differences at the cold side relative to warm side). You can compute efficiency of the power cycle based on the temperatures at the beginning of each step hence ? 1 ? , /? , will work with a small error as long as temperature change for each step is small. For a smaller approximation error in thermal efficiency, you could also try, ? 1 ? , ? , / ? , ? , . However, this will require finding a relationship between ? , and ? and solving 2 equations in 2 unknowns or having longer algebraic derivation to find both, which is possible. Former version is easier and later version has smaller error. You are free to choose either method but state which one you are using. You should repeat step by step solution until the temperatures of the high and low temperature reservoir reach the equilibrium conditions similar to Q1. Hint2: During the iterations temperature of the warm liquid will go down, temperature of the cold liquid will go up. At some point they will be close enough. You should check the convergence (equilibrium point) otherwise your iterations may continue forever. You need to check if the temperature difference between warm and cold liquids is smaller than what will result in at the end of the iterative step. Or you need to check it at the end of each step to make sure that cold liquid temperature does not increase above the warm liquid temperature. This will never happen in real life. But may happen in numerical iterations because of the approximation errors. You can experiment with larger temperature change or smaller temperature change for each step and see how your solution is affected. You will see if you make temperature step for each iteration large, you will have larger approximation errors. If you make it smaller, you will minimize errors, but you will need a lot of iterations. You can set your step size whatever makes the solution reasonable and explain how you made the decision. Q3: Repeat the Q2 assuming there are some irreversibilities associated with operation of heat engine and there are resistances to energy flow such that amount of heat passing through the components cause a temperature drop in the direction of heat. Figure 3 depicts an incremental step during this process. Assume FIGURE 3 there is a heat resistance (? 0.1 K/kJ) that causes a temperature drop (∆? ? δ? ) between the heat engine and the warm liquid (note that ∆? will change as δ? changes for each step). there is a heat resistance (? 0.1 K/kJ) that causes a temperature drop (∆? ? ?? ) between the heat engine and the cold liquid (note that ∆? will change as δ? changes for each step) thermal efficiency of the real heat engine is 80% of Carnot efficiency (note that real efficiency will change as the temperature of the warm and cold liquids change). In addition, real engine operates between warm liquid after temperature dropped (? |∆? | and cold liquid before temperature dropped (? ∆? (see Figure 3) Discuss the implications and effect of practical considerations on the entropy production and exergy destruction when you compare solution of this problem with problem 2.